In an `A.P.`, the `p^(th)` term is `1/p` and the `q^(th)` term is `1/p`. find the `(pq)^(th)` term of the `A.P.`

Explanation

Given , ` t_(p) = (1)/(q) and t_(q) = (1)/(2)`
` rArr a + (p-1) d = (1)/(q) ` …(i)
` rArr a + (p -1) d = (1)/(p) ` …(ii)
From (i)- (ii) , we get
` (p - 1-q + 1) d = (1)/(q) - (1)/(p) `
` rArr (p - q)d = (p-q)/(pq) `
` rArr = (1)/(pq)`
from (i) , ` a + (p-1) (1)/(pq) = (1)/(q)`
` rArr a = (1)/(q) - (1)/(q) + (1)/(pq) = (1)/(pq)`
` therefore a_(pq) = a+ (pq -1) d `
` = (1)/(pq) + (pq - 1) xx (1)/(pq)`
` (1)/(pq) (1 + pq - 1) = (pq)/(pq) = 1`